Let $R$ be the region enclosed by the polar curve $r(\theta)=\theta+\sin(\theta)$ where $\dfrac{\pi}{2}\leq \theta\leq \dfrac{3\pi}{4}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^{\scriptsize\dfrac{\pi}{4}}\left(\dfrac{1}{2}\theta^2+\theta\sin(\theta)+\dfrac{1}{2}\sin^2(\theta)\right)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{4}}\left(\dfrac{1}{2}\theta^2+\theta\sin(\theta)+\dfrac{1}{2}\sin^2(\theta)\right)d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{4}}\left(\theta^2+2\theta\sin(\theta)+\sin^2(\theta)\right)d\theta$ (Choice D) D $ \int_0^{\scriptsize\dfrac{\pi}{4}}\left(\theta^2+2\theta\sin(\theta)+\sin^2(\theta)\right)d\theta$
Explanation: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=\theta+\sin(\theta)}$, ${\alpha=\dfrac{\pi}{2}}$, and ${\beta=\dfrac{3\pi}{4}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{2}}}^{{\scriptsize\dfrac{3\pi}{4}}}\dfrac{1}{2}\left({\theta+\sin(\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{4}}\dfrac{1}{2}\left(\theta^2+2\theta\sin(\theta)+\sin^2(\theta)\right)d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{4}}\left(\dfrac{1}{2}\theta^2+\theta\sin(\theta)+\dfrac{1}{2}\sin^2(\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{4}}\left(\dfrac{1}{2}\theta^2+\theta\sin(\theta)+\dfrac{1}{2}\sin^2(\theta)\right)d\theta$